Could indeed be but I didn't factor a out of everything so to speak. I could've but I didn't. It checks out in line 3 as I move c across and divide both sides by a. Same result as if I had c/a in line 2.

(Original post by Ano9901whichone)
How would you solve this equation.
(x^2+2x+1)^2 -5(x^2+2x+1)+6=0. I tried expanding it all out it I couldn't do anything from there.

don't expand it out! That will take far more time than necessary.

If you look at this example it may help:

x^4 - 13x^2 + 36 = 0Now, at first it may not seem like it, but this is a quadratic 'in disguise'. You want to get this equation into the form of a quadratic (ax^2 + bx +c = 0) as you know how to solve quadratics.To do this, you want the first term to be to the power of 2, the second to the power of 1, and the last to be to the power of 0.

(x^2)^2 - 13(x^2)^1 + 36(x^2)^0

Looking at the rules of indices, you should see that this equation is equivalent to the original.

Now this looks complicated, so you can sub in a letter that would replace x^2 in your equation to clean it up a little.For instance, let y = x^2, and sub y into your new equation.

y^2 - 13x + 36 = 0

And now, this is a quadratic, something you know how to solve.Go ahead and factorise this:

(y - 4)(y - 9) = 0

And solving this, you get y = 4 or y = 9.BUT, this isn't your answer. If you remember from earlier, we said that y = x^2. So to work out what x is, you need to get the square roots of your answers, which will be +-2 and +-3

I know that may have been a long example, but can you see how that applies to your question?